LEAST-COST SCHEDULING METHOD OF PROJECT TIME CRASHING

Dr. Nida AzharDated : October 1, 2014

Definition

• A least cost schedule is one with optimum project duration such that to lengthen or shorten it would increase the project total cost.

TIME-COST RELATIONSHIPS

• Direct Costs – Direct costs, once a target schedule has been

developed, normally increase with a decrease in time.

• Indirect costs– Indirect costs, generally, vary directly with time,

i.e. a decrease in time causes a decrease in indirect costs, and vice versa.

Conditions governing the shortening of project duration (1/2)

• The shortening of project without increasing direct costs will increase the rate of capital turnover, and will reduce the indirect costs, i.e. those that vary directly with time. This may be the case when the technique to perform an activity is revised and replaced i.e. a more cost-effective technique is used which also provides a reduced project duration.

– By re-defining activity links (revised time planning)– By using more efficient construction methods– By improving resource productivity / using more efficient

resources– By optimizing available resources (revised resource & cost

planning)

Conditions governing the shortening of project duration (1/2)

• Increased direct costs due to crashed action may result in saving time, thereby increasing capital turnover, and possibly decreasing total costs as well, if the cost of the crash action is less than the incidence of indirect costs for the time saved. This is normally the case and we will consider this case to illustrate least-cost scheduling.

Cost Slope

• Normal time (DN) and normal cost (CN) are the values achieved when the activity is performed at minimum cost. Crash time (DC) and crash cost (CC) are the values for performing the activity in minimum time.

Cost slope = (CC – CN)/(DN – DC)

Method Approach

• The prime purpose of time-cost tradeoff study of a network is to reduce the project duration to the most economical duration, which is the point where the total cost curve is a minimum.

Method Assumptions

• The planned duration of an activity can be any whole day value between and including the normal and crash durations. For example, if A has a normal duration = 10 days and a crash duration = 7 days, then it has a possible duration = 7, 8, 9, or 10 days.

• Linear Compression i.e. the direct cost of an activity is linear between the normal and crash direct costs. For example, if A’s cost for a 10-day duration = Rs. 1700 and for a 7-day duration = Rs. 2000, then A’s cost for an 8-day duration = Rs. 1900 and for a 9-day duration = Rs. 1800.

Method Steps

1. Develop the project network diagram and perform CPM calculations considering normal project conditions.

2. Identify critical activities (these are the activities that will be crashed).3. Among critical activities, select an activity which has least cost slope.4. Compress the selected activity. An activity can be compressed maximum up to

its crash limit. 5. Note that if, during compression of the selected activity, another path

becomes critical, select another activity from the newly generated critical path having least cost slope and compress these two activities simultaneously.

6. Compress all critical activities till their crash limits (or up to desired limit). 7. Note that if there is more than one critical path, and crash limit of one is fully

consumed, crashing cannot be continued further. 8. The result will be the optimum time-cost solution.

Example

Example

Example

Example

Example

Example2nd Compression

Next critical activity with least cost slope is A Cost Slope = 50Crash Limit = 2 days

Compress Activity A by 2 days

4 8 8 13

2 B4 2 E5

6 10 10 15 15 15

0 4 4 60 G0

A4 8 C2 14 15 15 15

0 4 12 14 0 F1

4 14 14 15

0 D10

4 14

Duration: 15 days

Total Cost = 2300 + 2 x 50 - 2 x 100)Cost = Rs. 2200

Example3rd Compression

Next critical activity with least cost slope is D Cost Slope = 60Crash Limit = 5 days

But note that Activities B and E have total float value of 2. I f we compress Activity D by 5 days, then Activities B & D will becomecritical activities with negative total floats.So, we can safely compress activity D by 2 days only

Compress Activity D by 2 days

4 8 8 13

0 B4 0 E5

4 8 8 13 13 13

0 4 4 60 G0

A4 6 C2 12 13 13 13

0 4 10 12 0 F1

4 12 12 13

0 D8

4 12Duration: 13 daysTotal Cost = 2200 + 2 x 60 - 2 x 100Cost = Rs. 2120

Example

Since, there are two critical paths nowA --> D --> F --> GA --> B --> E --> GWe have to compress them simultaneoulsyActivity D has cost slope of 60 and crash limit is 3From B & E, Activity E has least cost slope of 25 and crash limit is 4Lower of the two crash limits (3 and 4) defines the combined crash limitSo, Compress Activities D and E simultaneously by 3 days each

ExampleCompress Activities D and E simultaneously by 3 days

4 8 8 10

0 B4 0 E2

4 8 8 10 10 10

0 4 4 60 G0

A4 3 C2 9 10 10 10

0 4 7 9 0 F1

4 9 9 10

0 D5

4 9Duration: 10 daysTotal Cost = 2120 + 3 x 60 + 3 x 25 - 3 x 100Cost = Rs. 2075

Now Path A- D - F has alredy saturated (no more compression can be done on any activity in this path because all activities have reached their crash limits).Hence, this is the optimum solution.

Optimum Result